Higher June 2019 Paper 2 Q5
5.
(a) Expand and simplify \((x + 6y)(3x + 5y)\). [3]
(b) Factorise \(x^2 - 13x + 36\). [2]
(c) Solve \(w^2 + 7w - 18 = 0\). [3]
(d) Factorise \(y^2 - 121\). [1]
(e) You are given that:
- \(y = x^2 + bx + c\)
- \(y = 16\) when \(x = 0\)
- \(y = 0\) when \(x = -2\)
| Scheme | Marks | Comments |
|---|---|---|
| \(3x^2 + 18xy + 5xy + 30y^2\) | B2 | B1 for any 2 terms correct |
| \(3x^2 + 23xy + 30y^2\) | B1 | FT for equivalent level of difficulty, providing at least 3 terms to consider and like terms to collect |
| Scheme | Marks | Comments |
|---|---|---|
| \((x - 9)(x - 4)\) | B2 | B1 for \((x \ldots 9)(x \ldots 4)\) |
| Scheme | Marks | Comments |
|---|---|---|
| \((w + 9)(w - 2) = 0\) | B2 | B1 for \((w \ldots 9)(w \ldots 2)\) |
| \(w = -9\) with \(w = 2\) | B1 | STRICT FT from ‘their pair of brackets’ |
Alternative
| Scheme | Marks | Comments |
|---|---|---|
| \((w=)\ \{-7 \pm \sqrt{(7^2 - 4\times1\times-18)}\}/2\) | M1 | |
| \((=)\ (-7 \pm \sqrt{121})/2\) | A1 | |
| \(w = -9\) with \(w = 2\) | A1 |
Accept trial & improvement method only if both solutions are found correctly for B3
| Scheme | Marks | Comments |
|---|---|---|
| \((y - 11)(y + 11)\) | B1 |
| Scheme | Marks | Comments |
|---|---|---|
| \(c = 16\) | B1 | |
| \(0 = (-2)^2 + b \times -2 + c\) | M1 | Allow \(-2^2 + b\times-2 + c\) |
| \(0 = (-2)^2 + b \times -2 + 16\) | M1 | FT ‘their derived 16’. Do not allow \(-2^2 + b\times-2 + 16\) unless \(-2^2\) seen or implied later as 4 |
| \(b = 10\) | A1 | |
| (13) |